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Popular Calculus >

integral from 0 to pi of sin^3(xco)s^2x

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Solution

∫0π​sin3(xco)s2x

Solution

9c2o2s2(3πcocos3(πco)−9πcocos(πco)+sin3(πco)+6sin(πco))​
Solution steps
∫0π​sin3(xco)s2xdx
Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx=s2⋅∫0π​sin3(cox)xdx
Apply u-substitution
=s2⋅∫0coπ​c2o2usin3(u)​du
Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx=s2c2o21​⋅∫0coπ​usin3(u)du
Apply Integration By Parts
=s2c2o21​[u(−cos(u)+31​cos3(u))−∫−cos(u)+31​cos3(u)du]0πco​
∫−cos(u)+31​cos3(u)du=−32​sin(u)−91​sin3(u)
=s2c2o21​[u(−cos(u)+31​cos3(u))−(−32​sin(u)−91​sin3(u))]0πco​
Expand u(−cos(u)+31​cos3(u))−(−32​sin(u)−91​sin3(u)):−ucos(u)+31​ucos3(u)+32​sin(u)+91​sin3(u)
=s2c2o21​[−ucos(u)+31​ucos3(u)+32​sin(u)+91​sin3(u)]0πco​
Compute the boundaries:3π​cocos3(πco)−πcocos(πco)+91​sin3(πco)+32​sin(πco)
=s2c2o21​(3π​cocos3(πco)−πcocos(πco)+91​sin3(πco)+32​sin(πco))
Simplify=9c2o2s2(3πcocos3(πco)−9πcocos(πco)+sin3(πco)+6sin(πco))​

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